Word of the Day – Isomorphic

Let’s play a game. First, you need an opponent. Next, you each take turns selecting a number from the spinner below. The first person to collect three numbers that add up to 12 wins.

Here is an example game:

Person A chooses 4. Person B chooses 7. Person A chooses 6. Person B chooses 3. Person A chooses 2 and wins, since 4 + 6 + 2 = 12

Give the game a try yourself to get the feel of it. You should realize that it is a fairly challenging game. Often, you get a couple number at the start, but are then forced to ‘block’ your opponent. Personally, a large number of my games ended in a draw because no one was able to get the right numbers to sum to 12.

As in any game, we want to develop a winning strategy. To start, we could write down all the sets of three numbers that add up to 12:

  • 0 + 4 + 8
  • 0 + 5 + 7
  • 1 + 3 + 8
  • 1 + 4 + 7
  • 1 + 5 + 6
  • 2 + 3 + 7
  • 2 + 4 + 6
  • 3 + 4 + 5

Next, we notice that the number 4 shows up in four of the sums, while the other numbers only show up two or three times. Hence, choosing 4 first would be a reasonable move. From there, we could try to keep our list of sums handy and use it to our wit our opponent. However, this strategy is cumbersome at best, and does not seem very intuitive. Perhaps there is a better way to understand the game…

The study of mathematics involves discovering structure and pattern. Frequently, a mathematician may find two objects that appear to have a similar structure or form. In our case, we wish to find a second object that has a similar structure to our spinner game. Since we need to add up three of the nine total numbers, we could consider a 3 X 3 grid:

Now we need to place our nine numbers somewhere in the grid. Since we want winning combinations, we should arrange all nine numbers in the grid so that every row, column, and diagonal adds up to 12 (our 8 different sums we listed above). Since the middle square is involved in four different sums (down, across, left diagonal, right diagonal) we place our most common number, the number 4, in the middle:

After some guessing and checking, we find the following placements of numbers appear to work for the diagonal sums:

We fill in the remaining spaces with our remaining numbers, ensuring the sum of each column and row is always 12. Voila:

For those of you who are interested, this type of construction has a special name, a magic square.

Back to our game. Now that we have our spinner in a different form, we can try using the grid to help us determine some strategy. Let’s represent Player A’s choice with an X and Player B’s choice with an O. Here is the first game from the beginning of the post:

Hmm, I think I have seen this game before… When a mathematician finds two things that have the same structure or form, they use a special word. That word is Isomorphic, which originates from the Greek iso, meaning “equal,” and morphic, meaning “shape” or “form.”

Due to our investigation above, we can confidently state that the spinner game and Tic-tac-toe are Isomorphic. This revelation was surprising to me. The spinner game seemed complicated and required a lot of thinking. Tic-tac-toe, on the other hand, is a child’s game that I mastered a long time ago. The fact that these two games are isomorphic shows just how useful looking at a problem from a different perspective can be.


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The Monty Hall Problem

Step right up, step right up! Try your luck at our fabulous contest. Guess right and you win a new car! Guess wrong, and you have to take home a goat!

You’re given the choice of three doors: Behind one door is a car. Behind the other two, smelly goats! You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3. He shows you that door No. 3 has a goat. He then says to you, “Do you want to switch to door No. 2?” Is it to your advantage to switch your choice?

Many people want to stay with their choice. This is because their gut reaction told them to pick door No. 1 and there is no reason to switch now. In fact, some contestants may even be suspicious of the game show hosts intentions. How do they know he won’t trick them and have them switch to a door with a goat!? A mathematical analysis is needed if clear thinking is to prevail.

As in similar situations, we need to break down this problem in to cases. To simplify the analysis, we will assume the car is always behind Door No. 1. This is because picking any other door and then doing a case by case analysis would be symmetric no matter what the numberings.

Case 1 (Pick Door No. 1):

\begin{array}{|c|c|c|}\hline  \textbf{Door No. 1} & \textrm{Door No. 2} & \textrm{Door No. 3} \\ \hline  \textrm{Car} & \textrm{Goat} & \textrm{ Goat}  \\ \hline \end{array}

Suppose you picked the door with the car, Door No. 1. But you don’t know it yet. And worse, the game show host is now opens door No. 2 and shows you a goat. He suggests you switch to door No. 3. If you switch, you will get a goat and lose. If you stay, you win a new car! Hmm, sticking with your original choice seems ideal in this situation.

Case 2 (Pick Door No. 2):

\begin{array}{|c|c|c|}\hline  \textrm{Door No. 1} & \textbf{Door No. 2} & \textrm{Door No. 3} \\ \hline  \textrm{Car} & \textrm{Goat} & \textrm{ Goat}  \\ \hline \end{array}

Now you picked a door with a goat, yuk! The host opens door No. 3 and shows you a goat. If you switch to door No. 1, you win the car. If you stay, you “win” the goat, not a good plan.

Case 3 (Pick Door No. 3):

\begin{array}{|c|c|c|}\hline  \textrm{Door No. 1} & \textrm{Door No. 2} & \textbf{Door No. 3} \\ \hline  \textrm{Car} & \textrm{Goat} & \textrm{ Goat}  \\ \hline \end{array}

Last case, you again have a door with a goat. The host shows you door No. 2 with a goat. If you switch to door No. 3 you win the car. If you stay, sadly, you will go home with the goat. Now let’s evaluate the strategies.

Say you are convinced that your gut is always right, and you never change your door. In case 1 you won. But your paranoia caused you to lose out big time in case 2 and 3. Hence, your chance of winning with the “stay” strategy is 1/3.

What if you are an indecisive person and you always change your mind (and your door) whenever possible. In case 1, you would change away from the car and get a goat. But in cases 2 and 3 your indecisiveness gave you the victory. Hence, your chance of winning with the “change” strategy is 2/3.

We can see that, overall, the better strategy is to change to the other door. This should be clear by the analysis above, and if you don’t believe me, try it for yourself next time you are on a game show.

However, if you would indulge me for just a moment, I want to muse on an interpretation of these results. If you pick any door at random, your chance of getting the car is 1/3. Next, the host opens a door to show you a goat. The math above states that, just by opening that door, the game has radically changed. If you stay with your door, you have a 1/3 chance of winning, just like before. But if you switch doors, suddenly your chance of winning jumps to 2/3!

But think about it logically. If there are only 2 doors left, and one has a goat and one has a car, shouldn’t your probability be 50-50, 1/2 chance either if you stay with your door or if you switch?

This is where the mystery of the problem comes in. Most people (wrongly) assume that since there are 2 doors left and 1 car they have a 50% chance of winning. So they see no reason to switch doors. In reality however, switching increases their chances of winning from 33% to 67%. But you protest, originally all 3 doors just had a 33% chance of winning, what in the world is going on?

The best I can explain it is this. When the game show host opened a door, he removed that door from the game. The 33% probability that that door was a winner had to go somewhere. And it did. This probability “collapsed” into the probability of the other door, giving the “switch” door a probability of 67% compared with your current “stay” door probability of 33%. Mind blowing. If you want the car, switch your door. If you want a goat, just say nope 🙂

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When it Doesn’t Distribute

One of the things I, and probably all math teachers, struggle with is teaching students when they can and cannot distribute. For example, students learn very early on in their mathematical education that:

3(4+8) = 3\times 4 + 3 \times 8 = 12 + 24 = 36

One reason we know this is valid is because we can calculate the answer using the rules of BEDMAS and obtain the same final answer:

3(4+8) = 3(12) = 36

This pattern extends to variables as well. Often it takes students a few tries, but eventually they become proficient in questions like:

5(x+2) = 5x + 10

However, the problem begins in grade 10. Students begin to encounter expressions like (x+3)^2. They believe that the same pattern should hold. And so, they dutifully write:

(x+3)^2 = x^2 + 3^2 = x^2+9

Next, they apply this pattern of distributing incorrectly to square roots:

\sqrt{9+16} = \sqrt{9} + \sqrt{16} = 3 + 4 = 7

Where clearly the answer should be:

\sqrt{9+16} = \sqrt{25} = 5

And once grade 11 and 12 hit, the number of bad distributing examples proliferates!

(x+2)^3 = x^3 + 2^3 = x^2+8

\sin (x+y) = \sin x + \sin y

\log{(x+10)} = \log x + \log 10

5(xy) = 5x5y

Ultimately, I think we as math teachers are to blame for the problem. Since distributing is so natural for us, we gloss over how unique and remarkable it actually is. As such, we give students a cavalier attitude towards distributing and they happily apply it in every context they encounter. As the above list suggests, distributing is very rare and only applies in a vary narrow set of situations. When teaching the distributive property, we should do a better job demonstrating just how unusual this property is.



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Complex Connections

The other day I was investigating a problem involving complex numbers when I was hit by a flash of insight. Here is the problem as it was originally stated:

If z is a complex number such that |z| = 1, compute:

|1-z|^2 + |1+z|^2

Initially, the problem seemed very dry. My first solution involved multiplying out a bunch of terms, combining like terms, and arriving at an answer of 4. However, as I was drawing a diagram of the problem, something amazing happened.

I realized that the expression |1-z| could be interpreted as the distance from the point (1,0) to the number . But the constraint that |z| = 1,  meant z had to lie on a circle of radius 1 centered at the origin.

Further, the expression |1+z| = |1-(-z)|  could be interpreted as the distance from the point (1,0) to the number -z, which would be directly opposite the original z .

From here, all I had to do was connect the dots and observe the solution. I had formed a right-triangle! If you want to play around with the diagram, feel free to click on the interactive link: https://ggbm.at/bJucvN5B

Hence, the expression: |1-z|^2 + |1+z|^2 , could be rewritten as:

(\textrm{side}_1)^2 + (\textrm{side}_2)^2

And using Pythagoras’s theorem, we get:

a^2 + b^2 = c^2

Now the hypotenuse is the diameter of the circle. Since the circle has radius 1, the diameter is 2, and:

|1-z|^2 + |1+z|^2 = 2^2 = 4

The question ultimately boiled down to using Pythagoras’s theorem. I continue to be amazed at how many times Pythagoras’ theorem is used. A simple relationship, taught to students in grade 7 or 8, continues to be relevant when evaluating complex valued expressions. Remarkable.

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Multiple Choice vs Matching Part 2

In the last post, we discovered that random guessing results in the same average score on both a matching question and a set of multiple choice questions. However, I still feel as though there is a difference between the two types of questions. Personally, the fact that it is impossible to score a 3 on the matching question piques my interest. To investigate the problem, we can use some of the results from the previous post.

Recall, we had the following table for the matching problem:

\begin{array}{|c|c|c|c|}\hline 0 \textrm{ Points} & 1 \textrm{ Point} & 2 \textrm{ Points}& 4 \textrm{ Points}\\ \hline 9 \textrm{ Ways} & 8 \textrm{ Ways} & 6 \textrm{ Ways} & 1 \textrm{ Way} \\ \hline  \end{array}

We can convert the ‘ways’ into probabilities by dividing by the total number of ways, 24. This gives:

\begin{array}{|c|c|c|c|c|}\hline 0 \textrm{ Points} & 1 \textrm{ Point} & 2 \textrm{ Points} & 3 \textrm{ Points}& 4 \textrm{ Points}\\ \hline 37.5\% & 33.3\%  & 25 \% & 0 \% & 4.2\% \\ \hline  \end{array}

Next, we need to compute the same table for the multiple choice situation. The math is simple, if a bit long:

\begin{array}{|c|c|c|c|c|}\hline 0 \textrm{ Points} & 1 \textrm{ Point} & 2 \textrm{ Points} & 3 \textrm{ Points}& 4 \textrm{ Points}\\ \hline 31.6\% & 42.2\%  & 21.1 \% & 4.7 \% & 0.4\% \\ \hline  \end{array}

That table is strange indeed. Comparing the two types of questions, nothing looks the same between them! Starting on the left, we see that with the matching question, you are around 6% more likely to score no points when compared to the multiple choice questions. That’s 6 more students out of 100 who would score no marks! Then, on the far right, you are 10 times more likely to score 4 points (full marks) on the matching than on the multiple choice! If you had a class of 100, 4 would score full marks on the matching while none would score full marks on the multiple choice. That hardly seems fair.

In fact, it feels like the matching question is ‘all or nothing.’ While you might score 1 point on average (see the previous post), you are more likely to score no points or full points than if you had been guessing on a set of multiple choice questions.

In statistics, we quantify this variability in the outcome using the standard deviation. Using this language, we would say our matching question has a higher standard deviation than our multiple choice question, even though they have the same average.

As an analogy, consider a room of big NFL players and skinny mathematicians. The average weight of a human in the room is probably around 200 pounds. However, there will be a huge variability in that weight; with some people weighing in around 275 and others around 150. This means our room would have a high standard deviation.

If instead, you had a room consisting of only 5 foot 10-inch-tall males with an average build, the average weight might still be around 200 pounds. But most of these people would be close to the average weight. This means our second room would have a low standard deviation.

So which question type (or room) is fairer? It really comes down to preference. While having question types with low variability might seem ideal, this is not set in stone. We appreciate the variability in the human population. If everyone was basically the same size, football wouldn’t be nearly as interesting. In fact, taking the low variability idea to the extreme, we could create a test that consisted entirely of true-false questions. While this test might seem fairer, I doubt many students would enjoy it. I for one, appreciate variability in my test questions.

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Multiple Choice vs Matching Part 1

I was making a test the other day and it got me thinking about question types. Which is fairer for students: a multiple-choice question, or a matching question? Of course, if a student knows the correct answer, then either question is equally fair since they will get full marks regardless of the question type. Instead, I was imagining a student who is uncertain of the correct answer and might need to make a guess. For example, consider the following 4 multiple choice questions:

Which of the following is a red fruit?

a)      Blueberry

b)      Banana

c)      Strawberry

d)      Orange

Which of the following is a yellow fruit?

a)      Blueberry

b)      Banana

c)      Strawberry

d)      Orange

Which of the following is a blue fruit?

a)      Blueberry

b)      Banana

c)      Strawberry

d)      Orange

Which of the following is an orange fruit?

a)      Blueberry

b)      Banana

c)      Strawberry

d)      Orange

Instead of having a ridiculous number of multiple choice questions, we could condense the above question set into a single matching question:

Match the following fruits to their colours:

A.      Blueberry

B.      Banana

C.      Strawberry

D.     Orange

1.      Red

2.      Orange

3.      Blue

4.      Yellow

Personally, I prefer the matching question; it feels clean and quick. Since most of you know your fruit colours, this isn’t a particularly difficult task. However, imagine if you had to take a guess on the colour of a blueberry. The problem is, if you incorrectly identify the blueberry as red, then no matter what you pick for the strawberry, your response will be incorrect.

Further, if you guess that strawberry matches with yellow, now the banana cannot pair with the correct colour. This cascading failure made me think that the matching question type might be setting students up for failure. Instead of going with my gut, I decided to investigate using math.

In the follow analysis, we consider the worst case scenario: a student who knows nothing about fruit colours and guesses at random. First, we consider the multiple choice question set. Since each question has 4 options, the student has a \frac{1}{4} probability of getting a question correct. Given that there are 4 questions, this means that the student on average will get one question correct. Thus, they will earn 1 mark. This makes intuitive sense to those of us who have guessed on multiple choice tests before.

Second, we need to analyze our matching question. Unfortunately, it is unclear how to proceed. The student has a \frac{1}{4} probability of matching the blueberry with the colour blue. But what if the student matches the blueberry with the colour red? This will affect the match on the strawberry, and so on. To make our analysis easier, we will consider a visual representation of the matching problem:

A possible student answer is:

In this example, the student scored 2 points because they had 2 correct matches. Now all we have to do is draw all possible matches and determine how many points each match gives. How many possible answers do you think there will be? We can start with answers that earn 4 points:

Not surprisingly, there is only one way to get everything correct.

Next, how many answers earn 3 points? If you think about it for a bit, you will realize that there are none. It is impossible for a student to score 3 points. How about earning 2 points?


We find 6 possible ways of earning 2 points. I will leave it as an exercise for the reader to confirm that there are 8 possible ways of earning 1 point. There are 4 ways of choosing A, then 3 remaining ways of choosing B (since A has already been matched to something), 2 ways of choosing C, and 1 way of choosing D, which gives 4 \times 3 \times 2 \times 1 = 24 possible ways of answering the matching question. Since we have found 1 + 6 + 8 = 15 ways so far, this means there are 24 - 15 =9 possible ways of answering and earning 0 points. Summarizing in a table:

\begin{array}{|c|c|c|c|}\hline 0 \textrm{ Points} & 1 \textrm{ Point} & 2 \textrm{ Points}& 4 \textrm{ Points}\\ \hline 9 \textrm{ Ways} & 8 \textrm{ Ways} & 6 \textrm{ Ways} & 1 \textrm{ Way} \\ \hline \end{array}

We calculate the average score by multiplying each number of points by the number of ways and then dividing by the total number of ways:
\frac{0 \times 9 + 1 \times 8 + 2 \times 6 + 4\times 1}{24} = \frac{24}{24} = 1
Wow, we get the same result as the multiple choice set! According to the above calculations, if you guess randomly on a matching question, you will get an average of one correct answer.
So which setup is fairer? Apparently, they are equal if we are measuring equality by the average score by guessing randomly. As a teacher, this gives me confidence that including matching questions on my tests is not harsh or unfair. However, the fact that it is impossible for a student to score a 3 on a matching question does raise further questions…

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What is multiplication really?

Recently, I took a course on how mathematics is taught to younger children. Throughout the course, the professor continued to hammer home the concept that multiplication is making groups. For example, 3 \times 4 means making 3 groups of 4:

While this representation is indeed accurate, I realized that I no longer thought about multiplication in this way. Somewhere in my education, I had left the ‘groups of’ model behind. In my applications of mathematics (both as a teacher and a student), I often encounter statements like: 2.7 \times 4. This statement makes no sense using the ‘groups of’ model. How can you have 2.7 groups of 4? Heck, how can you have 2.7 groups of anything?

However, to my mind, this situation is simple. I think of multiplication as scaling or stretching something. For example, 3 \times 4 means imagining a line of length 4 and then stretching it to be 3 times longer:

So when a question like 2.7 \times 4 appears, I just imagine stretching the line of length 4 a bit less than 3 full times.

The more I thought about the difference, the more it appeared to me to be based on two different definitions of multiplication. The ‘groups of’ idea originates with multiplication being defined as repeated addition. In our example, 3 \times 4 means 4+4+4 which we visualize as 3 groups of 4. The ‘scaling’ idea originate with vectors in a vector space where multiplication is defined much more abstractly.

When I reflect on my education, I realize that no one told me that we were switching from one idea to the next. Since I loved math, this probably didn’t bother me. I don’t remember any frustration understanding multiplication. However, for some of my students who are less mathematically inclined, I could see this switch producing real hardship. If they have mastered the ‘groups of’ concept, but are constantly being asked to calculate 6.2 \times 4.7, it is understandable why they would have difficulty interpreting their answer.

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